What is the general solution for -2sec(θ)+2sqrt(3)=0,0<= θ<= 360 ?

The general solution for -2sec(θ)+2sqrt(3)=0,0<= θ<= 360 is θ=0.95531…,θ=360-0.95531…

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-2sec(θ)+2sqrt(3)=0,0<= θ<= 360

- 2 sec (θ) + 23 = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 0.95531 \dots, θ = 36 0^{\circ} - 0.95531 \dots

Radians

θ = 0.95531 \dots, θ = 2 π - 0.95531 \dots

Solution steps

- 2 sec (θ) + 23 = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

Move

23

to the right side

- 2 sec (θ) + 23 = 0

Subtract

23

from both sides

- 2 sec (θ) + 23 - 23 = 0 - 23

Simplify

- 2 sec (θ) = - 23

- 2 sec (θ) = - 23

Divide both sides by

- 2

- 2 sec (θ) = - 23

Divide both sides by

- 2

\frac{- 2 sec ( θ )}{- 2} = \frac{- 2 3}{- 2}

Simplify

\frac{- 2 sec ( θ )}{- 2} = \frac{- 2 3}{- 2}

Simplify

\frac{- 2 sec ( θ )}{- 2} : sec (θ)

\frac{- 2 sec ( θ )}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 sec ( θ )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sec (θ)

Simplify

\frac{- 2 3}{- 2} : 3

\frac{- 2 3}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3

sec (θ) = 3

sec (θ) = 3

sec (θ) = 3

Apply trig inverse properties

sec (θ) = 3

General solutions for

sec (θ) = 3

sec (x) = a \Rightarrow x = arcsec (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arcsec (a) + 36 0^{\circ} n

θ = arcsec (3) + 36 0^{\circ} n, θ = 36 0^{\circ} - arcsec (3) + 36 0^{\circ} n

θ = arcsec (3) + 36 0^{\circ} n, θ = 36 0^{\circ} - arcsec (3) + 36 0^{\circ} n

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = arcsec (3), θ = 36 0^{\circ} - arcsec (3)

Show solutions in decimal form

θ = 0.95531 \dots, θ = 36 0^{\circ} - 0.95531 \dots

cos (x + 1) = - \frac{2}{5}

4 cos (2 x) + 1 = 3

sinh^{2} (x) = 0

cos (x) csc^{2} (x) + 3 cos (x) = 7 cos (x)

sin (x) - cos (x) = 1, 0 \leq x < 2 π

What is the general solution for -2sec(θ)+2sqrt(3)=0,0<= θ<= 360 ?
The general solution for -2sec(θ)+2sqrt(3)=0,0<= θ<= 360 is θ=0.95531…,θ=360-0.95531…